10.3 Vector Valued Functions
Greg Kelly, Hanford High School, Richland, Washington
Any vector
v  a, b
can be written as a linear
combination of two standard unit vectors.
i  1,0
v  a, b
 a,0  0, b
 a 1,0  b 0,1
 ai  bj
j  0,1
The vector v is a linear combination
of the vectors i and j.
The scalar a is the horizontal
component of v and the scalar b is
the vertical component of v.

v  a, b
 a,0  0, b
Either of these is an acceptable
way to express the same vector
function.
 a 1,0  b 0,1
 ai  bj

We can describe the position of a moving particle by a
vector, r(t).
r t 
f t  i
g t  j
r t   f t  i  g t  j
or
r(t) = f (t), g (t)
If we separate r(t) into horizontal and vertical components,
we can express r(t) as a linear combination of standard
unit vectors i and j.

In three dimensions the component form becomes:
r t   f t  i  g t  j  h t  k
Given the position vector:
r(t) = f (t), g (t)
which we can also write as… r(t) = x(t), y(t)
The velocity vector would be:
The acceleration vector would be:
dx dy
v(t) =
,
dt dt
d 2x d 2 y
, 2
a(t) =
2
dt dt

Graph on the TI-83 using the parametric mode.
r t   t cos t  i  t sin t  j
t 0
Use this
window setting:
This is just 8p

Graph on the TI-89 using the parametric mode.
r t   t cos t  i  t sin t  j
t 0
Hitting zoom fit followed by zoom square will give us…

Most of the rules for the calculus of vectors are the same as
we have used, except:
Speed  v  t 
“Absolute value” means
“distance from the origin”.
And since this also tells us that speed is the magnitude
________
of velocity, we must use the
____________________.
Pythagorean theorem.
v t 
velocity vector

Direction 
v t 
speed

Most of the rules for the calculus of vectors are the same as
we have used, except:
Speed  v  t 
dx dy
v(t) =
,
dt dt
Since we know what the
components of v(t) are…
2
dx   dy 

Speed  v t     
 dt   dt 
2
v t 
velocity vector

Direction 
v t 
speed


 

r  t   2t 3  3t 2 i  t 3  12t j
a) Write the equation of the tangent where

 
t  1.

dr
v t  
 6t 2  6t i  3t 2  12 j
dt
At t  1 :
position:
tangent:
v  1  12i  9j
r  1  5i 11j
 5,11
slope:
y  y1  m  x  x1 
3
y  11    x  5 
4
9
3

12
4
3
29
y  x
4
4


 

r  t   2t 3  3t 2 i  t 3  12t j

 

dr
v t  
 6t 2  6t i  3t 2  12 j
dt
b) Find the coordinates of each point on the path where
the horizontal component of the velocity is 0.
2
6
t
 6t .
The horizontal component of the velocity is
6t  6t  0
2
t t  0
r  0  0i  0j
 0, 0 
2
t  t 1  0
t  0, 1
r 1   2  3 i  1 12 j
r 1  1i 11j
 1, 11


 

dr
v t  
 6t 2  6t i  3t 2  12 j
dt
The velocity vector is often called the tangent vector.

Now let’s try an initial value problem:
 1 
v =
 i  2t  j
 t 1 
r(0) = i – 2j
Find the vector function r (t)
 dt  
r =
i
 t 1
 2t dt  j


 ln(t  1)  C1 i  t 2  C2 j


r(0) = i – 2j  ln(0  1)  C1 i  02  C2 j
C1  1
C2  2


r  ln(t  1)  1 i  t 2  2 j
p